Respuesta :
Answer:
a) Probability that haul time will be at least 10 min = P(X ≥ 10) ≈ P(X > 10) = 0.0455
b) Probability that haul time be exceed 15 min = P(X > 15) = 0.000
c) Probability that haul time will be between 8 and 10 min = P(8 < X < 10) = 0.6460
d) The value of c is such that 98% of all haul times are in the interval from (8.46 - c) to (8.46 + c)
c = 2.12
e) If four haul times are independently selected, the probability that at least one of them exceeds 10 min = 0.1700
Step-by-step explanation:
This is a normal distribution problem with
Mean = μ = 8.46 min
Standard deviation = σ = 0.913 min
a) Probability that haul time will be at least 10 min = P(X ≥ 10)
We first normalize/standardize 10 minutes
The standardized score for any value is the value minus the mean then divided by the standard deviation.
z = (x - μ)/σ = (10 - 8.46)/0.913 = 1.69
To determine the required probability
P(X ≥ 10) = P(z ≥ 1.69)
We'll use data from the normal distribution table for these probabilities
P(X ≥ 10) = P(z ≥ 1.69) = 1 - (z < 1.69)
= 1 - 0.95449 = 0.04551
The probability that the haul time will exceed 10 min is approximately the same as the probability that the haul time will be at least 10 mins = 0.0455
b) Probability that haul time will exceed 15 min = P(X > 15)
We first normalize 15 minutes.
z = (x - μ)/σ = (15 - 8.46)/0.913 = 7.16
To determine the required probability
P(X > 15) = P(z > 7.16)
We'll use data from the normal distribution table for these probabilities
P(X > 15) = P(z > 7.16) = 1 - (z ≤ 7.16)
= 1 - 1.000 = 0.000
c) Probability that haul time will be between 8 and 10 min = P(8 < X < 10)
We normalize or standardize 8 and 10 minutes
For 8 minutes
z = (x - μ)/σ = (8 - 8.46)/0.913 = -0.50
For 10 minutes
z = (x - μ)/σ = (10 - 8.46)/0.913 = 1.69
The required probability
P(8 < X < 10) = P(-0.50 < z < 1.69)
We'll use data from the normal distribution table for these probabilities
P(8 < X < 10) = P(-0.50 < z < 1.69)
= P(z < 1.69) - P(z < -0.50)
= 0.95449 - 0.30854
= 0.64595 = 0.6460 to 4 d.p.
d) What value c is such that 98% of all haul times are in the interval from (8.46 - c) to (8.46 + c)?
98% of the haul times in the middle of the distribution will have a lower limit greater than only the bottom 1% of the distribution and the upper limit will be lesser than the top 1% of the distribution but greater than 99% of fhe distribution.
Let the lower limit be x'
Let the upper limit be x"
P(x' < X < x") = 0.98
P(X < x') = 0.01
P(X < x") = 0.99
Let the corresponding z-scores for the lower and upper limit be z' and z"
P(X < x') = P(z < z') = 0.01
P(X < x") = P(z < z") = 0.99
Using the normal distribution tables
z' = -2.326
z" = 2.326
z' = (x' - μ)/σ
-2.326 = (x' - 8.46)/0.913
x' = (-2.326×0.913) + 8.46 = -2.123638 + 8.46 = 6.336362 = 6.34
z" = (x" - μ)/σ
2.326 = (x" - 8.46)/0.913
x" = (2.326×0.913) + 8.46 = 2.123638 + 8.46 = 10.583638 = 10.58
Therefore, P(6.34 < X < 10.58) = 98%
8.46 - c = 6.34
8.46 + c = 10.58
c = 2.12
e) If four haul times are independently selected, what is the probability that at least one of them exceeds 10 min?
This is a binomial distribution problem because:
- A binomial experiment is one in which the probability of success doesn't change with every run or number of trials. (4 haul times are independently selected)
- It usually consists of a number of runs/trials with only two possible outcomes, a success or a failure. (Only 4 haul times are selected)
- The outcome of each trial/run of a binomial experiment is independent of one another. (The probability that each haul time exceeds 10 minutes = 0.0455)
Probability that at least one of them exceeds 10 mins = P(X ≥ 1)
= P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
= 1 - P(X = 0)
Binomial distribution function is represented by
P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ
n = total number of sample spaces = 4 haul times are independently selected
x = Number of successes required = 0
p = probability of success = probability that each haul time exceeds 10 minutes = 0.0455
q = probability of failure = probability that each haul time does NOT exceeds 10 minutes = 1 - p = 1 - 0.0455 = 0.9545
P(X = 0) = ⁴C₀ (0.0455)⁰ (0.9545)⁴⁻⁰ = 0.83004900044
P(X ≥ 1) = 1 - P(X = 0)
= 1 - 0.83004900044 = 0.16995099956 = 0.1700
Hope this Helps!!!
